Two masses `m_(1)` and `m_(2)` at an infinite distance from each other are initially at rest, start interacting gravitationally. Find their velocity of approach when they are at a distance `r` apart.
Text Solution
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Since they move under mutual attraction and no external force acts on them, their momentum and energy are conserved. Therefore, `:.0=1/2M_(1)v_(1)^(2)+1/2M_(2)v_(2)^(2)-(GM_(1)M_(2))/s` It is zero because in the beginning, both kinetic energy and potential energy are zero. `0=M_(1)v_(1)+M_(2)v_(2)` Solving the equations `v_(1)^(2)=(2GM_(2)^(2))/(s(M_(1)+M_(2))` and `v_(2)^(2)=(2GM_(1)^(2))/(s(M_(1)+M_(2)))` `V` (velocity of approach) `=v_(1)-(-v_(2))=v_(1)+v_(2)` `=sqrt((2G(M_(1)+M_(2)))/s)`
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